Calculating bits required to store decimal number

Question

This is a homework question that I am stuck with      Consider unsigned integer representation  How many bits will be   required to store a decimal number containing    i  3 digits ii  4 digits iii  6 digits iv  n digits   I know that the range of the unsigned integer will be 0 to 2 n but I don t get how the number of bits required to represent a number depends upon it  Please help me out   Thanks in advance

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Keep dividing the number by 2 until you get a quotient of 0

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For a binary number of n digits the maximum decimal value it can hold will be  2 n - 1  and 2 n is the total permutations that can be generated using these many digits   Taking a case where you only want three digits  ie your case 1  We see that the requirements is   2 n - 1    999  Applying log to both sides   log 2 n - 1     log 999   log 2 n  - log 1     log 999   n   9 964  approx    Taking the ceil value of n since 9 964 can t be a valid number of digits  we get n   10

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The short answer is   int nBits   ceil log2 N      That s simply because pow 2  nBits  is slightly bigger than N

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Assuming that the question is asking what s the minimum bits required for you to store    3 digits number   My approach to this question would be     what s the maximum number of 3 digits number we need to store  Ans  999 what s the minimum amount of bits required for me to store this number    This problem can be solved this way by dividing 999 by 2 recursively  However  it s simpler to use the power of maths to help us  Essentially  we re solving n for the equation below   2 n   999 nlog2   log999 n   10   You ll need 10 bits to store 3 digit number    Use similar approach to solve the other subquestions   Hope this helps

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Ok to generalize the technique of how many bits you need to represent a number is done this way   You have R symbols for a representation and you want to know how many bits  solve this equation R 2 n or log2 R  n   Where n is the numbers of bits and R is the number of symbols for the representation      For the decimal number system R 9 so we solve 9 2 n   the answer is 3 17 bits per decimal digit   Thus a 3 digit number will need 9 51 bits or 10   A 1000 digit number needs 3170 bits

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The simplest answer would be to convert the required values to binary  and see how many bits are required for that value  However  the question asks how many bits for a decimal number of X digits  In this case  it seems like you have to choose the highest value with X digits  and then convert that number to binary    As a basic example  Let s assume we wanted to store a 1 digit base ten number  and wanted to know how many bits that would require  The largest 1 digit base ten number is 9  so we need to convert it to binary  This yields 1001  which has a total of 4 bits  This same example can be applied to a two digit number  with the max value being 99  which converts to  1100011   To solve for n digits  you probably need to solve the others and search for a pattern   To convert values to binary  you repeatedly divide by two until you get a quotient of 0  and all of your remainders will be 0 or 1   You then reverse the orders of your remainders to get the number in binary    Exampe  13 to binary    13 2    6 r 1 6 2     3 r 0 3 2     1 r 1 1 2     0 r 1   1101   8 1     4 1     2 0     1 1     Hope this helps out

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This one works   floor loge n    loge 2     1   To include negative numbers  you can add an extra bit to specify the sign   floor loge abs n     loge 2     2

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The largest number that can be represented by an n digit number in base b is bn - 1  Hence  the largest number that can be represented in N binary digits is 2N - 1  We need the smallest integer N such that      2N - 1   bn - 1     2N   bn   Taking the base 2 logarithm of both sides of the last expression gives      log2 2N   log2 bn     N   log2 bn     N   log bn   log 2   Since we want the smallest integer N that satisfies the last relation  to find N  find log bn   log 2 and take the ceiling   In the last expression  any base is fine for the logarithms  so long as both bases are the same  It is convenient here  since we are interested in the case where b   10  to use base 10 logarithms taking advantage of log1010n    n   For n   3      N    3   log10 2    10   For n   4      N    4   log10 2    14   For n   6      N    6   log10 2    20   And in general  for n decimal digits      N    n   log10 2

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The formula for the number of binary bits required to store n integers  for example  0 to n - 1  is   loge n    loge 2   and round up   For example  for values -128 to 127  signed byte  or 0 to 255  unsigned byte   the number of integers is 256  so n is 256  giving 8 from the above formula   For 0 to n  use n   1 in the above formula  there are n   1 integers    On your calculator  loge may just be labelled log or ln  natural logarithm

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There are a lot of answers here  but I ll add my approach since I found this post while working on the same problem   Starting with what we know here are the number from 0 to 16   Number           encoded in bits         minimum number of bits to encode 0                000000                  1 1                000001                  1 2                000010                  2 3                000011                  2 4                000100                  3 5                000101                  3 6                000110                  3 7                000111                  3 8                001000                  4 9                001001                  4 10               001010                  4 11               001011                  4 12               001100                  4 13               001101                  4 14               001110                  4 15               001111                  4 16               010000                  5   looking at the breaks  it shows this table  number  lt         number of bits 1               0 3               2 7               3 15              4   So  now how do we compute the pattern   Remember that log base 2  n    log base 10  n    log base 10  2   number    logb10  n    logb2  n    ceil logb2 n    1         0            0           0            special case  3         0 477        1 58        2 7         0 845        2 807       3   8         0 903        3           3            special case  15        1 176        3 91        4 16        1 204        4           4            special case  31        1 491        4 95        5 63        1 799        5 98        6   Now the desired result matching the first table   Notice how also some values are special cases   0 and any number which is a powers of 2   These values dont change when you apply ceiling so you know you need to add 1 to get the minimum bit field length     To account for the special cases add one to the input   The resulting code implemented in python is   from math import log from math import ceil def min num bits to encode number a number       a number a number 1    adjust by 1 for special cases        log of zero is undefined     if 0  a number          return 0     num bits   int ceil log a number 2       logbase2 is available     return  num bits

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let its required n bit then 2 n  base  digit and then take log and count no  for n

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Well  you just have to calculate the range for each case and find the lowest power of 2 that is higher than that range   For instance  in i   3 decimal digits -  10 3   1000 possible numbers so you have to find the lowest power of 2 that is higher than 1000  which in this case is 2 10   1024  10 bits    Edit  Basically you need to find the number of possible numbers with the number of digits you have and then find which number of digits  in the other base  in this case base 2  binary  has at least the same possible numbers as the one in decimal   To calculate the number of possibilities given the number of digits  possibilities base ndigits  So  if you have 3 digits in decimal  base 10  you have 10 3 1000 possibilities  Then you have to find a number of digits in binary  bits  base 2  so that the number of possibilities is at least 1000  which in this case is 2 10 1024  9 digits isn t enough because 2 9 512 which is less than 1000    If you generalize this  you have  2 nbits possibilities  lt   gt  nbits log2 possibilities   Which applied to i  gives  log2 1000  9 97 and since the number of bits has to be an integer  you have to round it up to 10

[decimal] Calculating bits required to store decimal number

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